Question Statement

Convert the following equations of circle into standard form and hence find their centre and radius:

(i) $(0,2),(2,0),(1,3)$

(ii) $5x^2+5y^2-2x+4y-27=0$

(iii) $4x^2+4y^2+2ax+by-a^2=0$

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Background and Explanation

The standard form of a circle is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ represents the centre and $r$ represents the radius. To convert from the general form $x^2+y^2+Dx+Ey+F=0$, we use the method of completing the square for both the $x$ and $y$ terms. This involves rearranging terms, ensuring the coefficients of $x^2$ and $y^2$ are equal to 1, and then adding the square of half the coefficient to each group.

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Solution

Part (i)

The given equation of the circle is:

$x^2+y^2-4x+6y-36=0$

First, we group the $x$ terms and $y$ terms together, moving the constant to the right-hand side:

$\begin{array}{rl}& x^2+y^2-4x+6y-36=0\\ \Rightarrow & (x^2-4x)+(y^2+6y)=36\end{array}$

Next, we complete the square for each group. For the $x$ terms: take half of $-4$ (which is $-2$) and square it to get $4$. For the $y$ terms: take half of $6$ (which is $3$) and square it to get $9$. We add these values to both sides of the equation:

$\Rightarrow (x^2-4x+4-4)+(y^2+6y+9-9)=36$

Now we rewrite the perfect square trinomials and simplify:

$\Rightarrow (x-2{)}^2-4+(y+3{)}^2-9=36$

Move the constants $-4$ and $-9$ to the right-hand side:

$\text{or}(x-2{)}^2+(y+3{)}^2=49$

This can also be written as:

$(x-2{)}^2+(y-(-3){)}^2=72$

Result: The standard form is $(x-2{)}^2+(y+3{)}^2=7^2$.
Centre: $(2,-3)$
Radius: $7$

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Part (ii)

The given equation of the circle is:

$5x^2+5y^2-2x+4y-27=0$

First, we make the coefficients of $x^2$ and $y^2$ equal to 1 by dividing the entire equation by 5:

$x^2+y^2-\frac{2}{5}x+\frac{4}{5}y-\frac{27}{5}=0$

Group the $x$ and $y$ terms and move the constant to the right:

$\Rightarrow \left(x^2-\frac{2}{5}x\right)+\left(y^2+\frac{4}{5}y\right)=\frac{27}{5}$

Complete the square:

• For $x$: half of $-\frac{2}{5}$ is $-\frac{1}{5}$, squared gives $\frac{1}{25}$
• For $y$: half of $\frac{4}{5}$ is $\frac{2}{5}$, squared gives $\frac{4}{25}$

Add these to both sides:

$\Rightarrow \left(x^2-\frac{2}{5}x+\frac{1}{25}-\frac{1}{25}\right)+\left(y^2+\frac{4}{5}y+\frac{4}{25}-\frac{4}{25}\right)=\frac{27}{5}$

Rewrite as perfect squares and simplify:

$\Rightarrow {\left(x-\frac{1}{5}\right)}^2-\frac{1}{25}+{\left(y+\frac{2}{5}\right)}^2-\frac{4}{25}=\frac{27}{5}$

Move the constants to the right and combine:

${\left(x-\frac{1}{5}\right)}^2+{\left(y+\frac{2}{5}\right)}^2=\frac{1}{25}+\frac{4}{25}+\frac{27}{5}$

$\Rightarrow {\left(x-\frac{1}{5}\right)}^2+{\left(y+\frac{2}{5}\right)}^2=\frac{28}{5}$

Or equivalently:

${\left(x-\frac{1}{5}\right)}^2+{\left(y-\frac{-2}{5}\right)}^2={\left(\sqrt{\frac{28}{5}}\right)}^2$

Result: The standard form is ${\left(x-\frac{1}{5}\right)}^2+{\left(y+\frac{2}{5}\right)}^2=\frac{28}{5}$.
Centre: $\left(\frac{1}{5},-\frac{2}{5}\right)$
Radius: $\sqrt{\frac{28}{5}}$

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Part (iii)

The given equation of the circle is:

$4x^2+4y^2+2ax+by-a^2=0$

Divide both sides by 4 to make the coefficients of $x^2$ and $y^2$ equal to 1:

$x^2+y^2+\frac{a}{2}x+\frac{b}{4}y-a^2=0$

Group the variables and move the constant to the right:

$\Rightarrow \left(x^2+\frac{a}{2}x\right)+\left(y^2+\frac{b}{4}y\right)=a^2$

Complete the square:

• For $x$: half of $\frac{a}{2}$ is $\frac{a}{4}$, squared gives $\frac{a^2}{16}$
• For $y$: half of $\frac{b}{4}$ is $\frac{b}{8}$, squared gives $\frac{b^2}{64}$

Add these to both sides:

$\Rightarrow \left(x^2+\frac{a}{2}x+\frac{a^2}{16}-\frac{a^2}{16}\right)+\left(y^2+\frac{b}{4}y+\frac{b^2}{64}-\frac{b^2}{64}\right)=a^2$

Rewrite as perfect squares:

$\Rightarrow {\left(x+\frac{a}{4}\right)}^2-\frac{a^2}{16}+{\left(y+\frac{b}{8}\right)}^2-\frac{b^2}{64}=a^2$

Move the constants to the right and combine terms:

$\Rightarrow {\left(x+\frac{a}{4}\right)}^2+{\left(y+\frac{b}{8}\right)}^2=\frac{b^2}{64}+\frac{a^2}{16}+a^2$

Simplifying the right-hand side:

${\left(x+\frac{a}{4}\right)}^2+{\left(y+\frac{b}{8}\right)}^2=\frac{b^2+60a^2}{64}$

Or written in the standard radius form:

${\left(x-\frac{-a}{4}\right)}^2+{\left(y-\frac{-b}{8}\right)}^2={\left(\frac{\sqrt{b^2+60a^2}}{8}\right)}^2$

Result: The standard form is ${\left(x+\frac{a}{4}\right)}^2+{\left(y+\frac{b}{8}\right)}^2=\frac{b^2+60a^2}{64}$.
Centre: $\left(-\frac{a}{4},-\frac{b}{8}\right)$
Radius: $\frac{\sqrt{b^2+60a^2}}{8}$

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Key Formulas or Methods Used

• Standard form of circle: $(x-h{)}^2+(y-k{)}^2=r^2$ where centre is $(h,k)$ and radius is $r$
• Completing the square: $x^2+Dx={\left(x+\frac{D}{2}\right)}^2-{\left(\frac{D}{2}\right)}^2$
• Normalizing coefficients: Dividing the entire equation by the coefficient of $x^2$ (and $y^2$) to make it equal to 1
• Grouping terms: Collecting $x$-terms and $y$-terms separately to apply completing the square method

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Summary of Steps

1. Normalize: If coefficients of $x^2$ and $y^2$ are not 1, divide the entire equation by that coefficient
2. Group: Rearrange to group all $x$ terms and all $y$ terms on the left, moving the constant to the right
3. Complete the square for $x$: Add ${\left(\frac{\text{coefficient of }x}{2}\right)}^2$ to both sides
4. Complete the square for $y$: Add ${\left(\frac{\text{coefficient of }y}{2}\right)}^2$ to both sides
5. Factor: Write the grouped terms as perfect square binomials: $(x-h{)}^2$ and $(y-k{)}^2$
6. Simplify: Combine all constant terms on the right side to get $r^2$
7. Identify: Read off the centre $(h,k)$ and radius $r=\sqrt{\text{RHS}}$